Triola Assignment F
Section 12-2 – Basic Skills and Concepts
1) What is one-way analysis of variance, and what is it used for? One-way ANOVA is a procedure for comparing three or more population means. The procedure uses one factor to separate the various samples and analyzed differences in the sample variance to determine if the means are equal or unequal. Check: OK
3) What is variance between samples? What if variance within samples?. Variance between samples addresses how the separate samples differ whereas the variance within samples addresses the variance in individual samples. Check: OK
5) We noted in the Chapter Problem that the weight of 1.34 kg appears to be an outlier. If we delete that value, the STATDISK results are as shown below. a) What is the null hypothesis?
Null hypothesis H0: 1 = 2 = 3 = 4
b) What is the alternative hypothesis? Alternative hypothesis H1: At least one of the means is significantly different
c) Identify the value of the test statistic. The value of the test statistic, F, is: 8.448
d) What is the critical value for a 0.05 significance level? The critical value is: 3.2874.
e) Identify the P-value. The P-value is: 0.0016.
f) Based on the preceding results, what do you conclude about the equality of the population means? Since the P-value is less than 0.05, you would reject the null hypothesis and conclude that the population means are not all equal.
g) Compare these results to those obtained with the weight of 1.34 kg included. Does the apparent outlier of 1.34 kg have much an effect on the results? Does the conclusion change? When the outlier is included in the data, the test statistic is F = 5.73135 and the P- value is 0.007. Although these value do change quite a bit, the results still suggest to reject the null hypothesis and conclude that the population means are not equal. Check: OK
Section 12-3 – Basic Skills and Concepts 1) What is two-way analysis of variance, and what is it used for? Two-way analysis of variance is essentially the same as one-way ANOVA except that two factors are considered in the process. This process is used to comparing populations means for three or more samples and for two factors. Check: OK
3) When conducting two-way analysis of variance, if there is an interaction between the two factors, why should we not continue with the separate tests for effects from the row and column factors? If an effect exists between the two factors, then you do not conduct the separate tests for effects from the factors because an interaction between the factors means that you cannot consider the effects of each factor individually. Check: OK
10) Use the Minitab display, which results from the scores listed in the accompanying table. The sample data are SAT scores on the verbal and math portions of the SAT-1 and are based on reported statistics from the College Board. Assume that SAT scores are not affected by an interaction between gender and the type of test. Is there sufficient evidence to support the claim that the type of test has an effect on SAT scores? The test statistic for type of test is: F= 0.58. The P-value is: 0.453. Since the P-value is not less that 0.05, we can conclude that the type of test does NOT have an effect on SAT scores. Check: OK
Unit 12 – Review Exercises
1) The Associated Insurance Institute sponsors studies of the effects of drinking on driving. IN one such study, three groups of adult men were randomly selected for an experiment designed to measure their blood alcohol levels after consuming five drinks. Member of groups A were test after 1 hour, members of group B were tested after 2 hours, and members of group C were tested after 4 hours. The results are given in the accompanying table; the Minitab display for these data is also shown. At the 0.05 significance level, test the claim that the three groups have the same mean level. The test statistic is: F = 46.90. The P-value is: 0.000. Since the P-value is less than 0.05, we can reject the null hypothesis and conclude that there is a difference in the means. Check: OK
3) Use the Minitab display, which results from the values listed in the accompanying table. The sample data are student estimates of the length of their classroom. The actual length of the classroom is 24 feet, 7.5 inches. Test the null hypothesis that the estimated lengths are not affected by an interaction between gender and major. The test statistics for the interaction is: F = 0.19. The P-value is: 0.832. Since the P- value is NOT less than 0.05, we fail to reject the null hypothesis. Therefore, the estimates are not affected by an interaction between gender and major. Check: OK
5) Use the Minitab display, which results from the values listed in the accompanying table. The sample data are student estimates of the length of their classroom. The actual length of the classroom is 24 feet, 7.5 inches. Assume that the estimated lengths are not affected by an interaction between gender and major. Is there sufficient evidence to support the claim that estimated length is affected by major? The test statistics for the effect of major is: F = 0.13. The P-value is: 0.876. Since the P- value is NOT less than 0.05, we fail to reject the null hypothesis. Therefore, the estimates are not affected by major. Check: OK
Unit 12 – Cumulative Review Exercises
1) Refer to the number of years that U.S. presidents and popes and British monarchs lived after their inauguration, election, or coronation. The data are listed in the table for Review Exercise 8. a) Find the mean for each of the three groups. 557 Presidents x 15.054054 15.054 37 315 Popes x 13.125 24 318 Monarchs x 22.714285 22.714 Check: OK. 14 b) Find the standard deviation for each of the three groups. Presidents x 2 102 292 ... 232 11603 2 x (10 29 ... 23) 2 5572 310249 2 n x 2 x 37(11603) 310249 119062 s nn 1 3737 1 1332 89.38588589 9.454410922 9.454 Popes x 2 22 92 ... 262 5981 2 x (2 9 ... 26) 2 3152 99225 2 n x 2 x 24(5981) 99225 44319 s nn 1 2424 1 552 80.28804348 8.960359562 8.960
Monarchs x 2 17 2 62 ... 152 11722 2 x (17 6 ... 15) 2 3182 101124 2 n x 2 x 14(11722) 101124 62984 s nn 1 1414 1 182 346.0659341 18.60284747 18.603 Check: OK. c) Test the claim that there is a difference between the mean for presidents and the mean for British monarchs.
Null hypothesis H0: 1 = 2
Alternative hypothesis H1: 1 2 Test statistic x x 15.054 22.714 0 7.66 7.66 t 1 2 1 2 s 2 s 2 9.4542 18.6032 2.415624 24.7194 27.1350254 1 2 n1 n2 37 14 7.66 1.470495343 1.470 5.209129044 Critical Value df 37 1 36 or df 14 1 13 Use 13 as the df since it is smaller. For an α = 0.05 in a two-tailed test, the critical value is: 2.160 Conclusion About Null Since –1.470 is NOT in the critical region, the conclusion is to fail to reject the null hypothesis. Final Conclusion Since we failed to reject the null hypothesis, is appears that there is not a difference between the mean longevity values for presidents and monarchs. Check: OK. d) Use the longevity times for presidents and determine whether they appear to be come from a population having a normal distribution. Explain why the distribution does or does not appear to be normal. The graph does appear to be somewhat normal. The values that fall between 0 and 5 are a little different; otherwise, the graph is close to normal. Check: OK.
Histogram of President Longevity
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Mean =15.05 Std. Dev. =9.454 N =37 0 0 10 20 30 40 president e) Use the longevity times for presidents and construct a 95% confidence interval estimated of the population mean. 1 0.95 0.05 t 2.028 s 9.454 n 37 s 9.454 9.454 E t / 2 2.028 2.028 2.0281.554228026 3.151974437 3.152 n 37 6.08276253 x 15.054 x E x E 15.054 3.152 15.054 3.152 11.902 18.206 Check: OK
3) Some couples have genetic characteristics configured so that one-quarter of all their offspring have blue eyes. A study is conducted of 100 couples believed to have those characteristics, with the result that 19 of their 100 offspring have blue eyes. Assuming that one-quarter of all offspring have blue eyes, estimate the probability that among 100 offspring, 19 or fewer have blue eyes. Based on that probability, does it seem that the one-quarter rate is wrong? Why or why not? 1 3 np 100 25 nq 100 75 This confirms that np and nq are greater than 5. 4 4 1 100 25 4 1 3 npq 100 18.75 4.330127019 4.330 4 4 Since this is discrete set of values and you want to include 19 in the probability, you need to use x = 19.5. x 19.5 25 5.5 z 1.270170592 1.270 4.330 4.330 The probability for the z-value of –1.27 is 0.1020. Based on this probability, there is not sufficient evidence to assume that 0.25 is wrong (i.e. 0.1020 in NOT less than 0.05) Check: OK. I forgot to adjust 19 to 19.5 since it was a discrete situation. Otherwise everything was okay.
Section 13-2 – Basic Skills and Concepts
1) Why is the sign test considered to be a “nonparametric” test or a “distribution-free” test? The sign test is “nonparametric” because the sample does not have to come from a population that aligns to a distribution (i.e. normal). Check: OK
3) You have been given the task of testing the claim that a method of gender selection has the effect of increasing the likelihood that a baby will be a girl, and sample data consist of 20 girls among 80 newborn babies. Without applying the formal sign test procedure, what do you conclude about the claim? Why? Given the situation, you would assume that 40 of babies (0.5) would be girls solely on chance. Since 0.25 (the proportion of girls in the samples) is less than 0.5, you would fail to reject the null before the test is even conducted. Check: OK
5) Assume matched pairs of data result in the given number of signs when the value of the second variable is subtracted from the corresponding values of the first variable. Use the sign test with a 0.05 significance level to test the null hypothesis of no difference. Positive signs: 15 Negative signs: 4 Ties: 1
The null hypothesis is: H0: There is no difference.
The alternative hypothesis is: H1: There is a difference (the median of differences 0) The significance level is: = 0.05 Use the nonparametric sign test. The test statistic is: x = 4 (the smaller of the two counts). N = 19 since ties are discarded. This is a two-tailed test and the critical value (from Table A-7) is: 4. Since the test statistic is less than or equal to 4, we reject the null hypothesis and conclude that there is a difference in the median values. Check: OK
10) In 1908, William Gosset published an article. He included the data listed below for yields from two different types of seed that were used on adjacent plots of land. The listed values are the yields of straw in cwt per acre, where cwt represents 100 lb. Using a 0.05 significance level, test the claim that there is no difference between the yields from the two types of seed. Does it appear that either type of seed is better? Reg 19.25 22.75 23 23 22.5 19.75 24.5 15.5 18 14.25 17 Kiln 25 24 24 28 22.5 19.5 22.25 16 17.25 15.75 17.25 – – – – 0 + + – + – – Positive signs: 3 Negative signs: 7 Ties: 1
The null hypothesis is: H0: There is no difference.
The alternative hypothesis is: H1: There is a difference (the median of differences 0) The significance level is: = 0.05 Use the nonparametric sign test. The test statistic is: x = 3 (the smaller of the two counts). N = 10 since ties are discarded. This is a two-tailed test and the critical value (from Table A-7) is: 1. Since the test statistic is NOT less than or equal to 1, we fail to reject the null hypothesis and conclude that there is no difference in the yields from regular and kiln-dried seeds. Check: OK
Section 13-3 – Basic Skills and Concepts
1) Why would we use the Wilcoxon signed-ranks test for matched pairs instead of the methods presented in Section 9-4? Wilcoxon signed-ranks test is used for cases when the sample comes form a population that does not align to a particular distribution (this is a nonparametric test). Check: OK
3) Given sample data consisting of matched pairs, we can compare the values using the sign test or the Wilcoxon signed-ranks test. What important advantages does the Wilcoxon test have over the sign test? The Wilcoxon signed-ranks test takes into account the magnitude of the differences whereas the sign test does not. Therefore, the Wilcoxon test tends to results in more accurate conclusions about the situation. Check: OK
5) Refer to the given paired sample data and use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with a median equal to zero. Use a 0.05 significance level. X 60 55 89 92 78 84 93 87 Y 35 27 47 44 39 48 51 54 Difference 25 28 42 48 39 36 42 33 Rank 1 2 6.5 8 5 4 6.5 3 Signed Rank 1 2 6.5 8 5 4 6.5 3
The null hypothesis is: H0: The matched pairs have differences that come from a population with a median equal to zero. The alternative hypothesis is: H1: The match pairs have differences that come from a population with a non-zero median. The significance level is: = 0.05 The sum of negative ranks: 0 The sum of positive ranks: 36 The test statistic is: T = 0 (the smaller of the two sums) n is 8. Because n = 8, we have n 30, so we use the test statistic of T = 0. This is a two-tailed test and the critical value (from Table A-8) is: 4. Since the test statistic is less than or equal to 4, we reject the null hypothesis and conclude that the matched pairs come from a population with a non-zero median. Check: OK
7) Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and the results are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with a median equal to zero. Use a 0.05 significance level. Friday (6th) 9 6 11 11 3 5 Friday (13th) 13 12 14 10 4 12 Difference –4 –6 –3 1 –1 –7 Rank 4 5 3 1.5 1.5 6 Signed Rank –4 –5 –3 1.5 –1.5 –6
The null hypothesis is: H0: The matched pairs have differences that come from a population with a median equal to zero. The alternative hypothesis is: H1: The match pairs have differences that come from a population with a non-zero median. The significance level is: = 0.05 The sum of negative ranks: 19.5 The sum of positive ranks: 1.5 The test statistic is: T = 1.5 (the smaller of the two sums) n is 6. Because n = 6, we have n 30, so we use the test statistic of T = 1.5. This is a two-tailed test and the critical value (from Table A-8) is: 1. Since the test statistic is NOT less than or equal to 1, we fail to reject the null hypothesis and conclude that the matched pairs come from a population with a median equal to zero. Friday the 13th does not affect the number of hospital admissions. Check: OK
Section 13-4 – Basic Skills and Concepts
1) What is the most fundamental difference between the Wilcoxon signed-ranks test and the Wilcoxon rank-sum test? For the Wilcoxon signed-ranks test, the data must be matched or paired. For the Wilcoxon rank-sum test, the sample are independent. Check: OK
3) The Wilcoxon rank-sum test and the methods of hypothesis testing described in Section 9-3 both apply to two independent samples. What advantage does the Wilcoxon rank-sum test have over the test described in Section 9-3? The Wilcoxon rank-sum test does not require normally distributed populations. Check: OK
5) Use a 0.05 significance level with the methods of this section to identify the rank sums R1 and
R2, R, σR, the test statistic z, the critical z values, and then state the conclusion about the claim of equal medians. Sample 1 2 7 10 16 20 22 23 26 27 30 33 RANK 1 4 5 8 9 10 11 12 13 15 16 Sample 2 3 4 11 14 28 35 40 46 47 52 53 60 RANK 2 3 6 7 14 17 18 19 20 21 22 23
R1: 1 + 4 + 5 + 8 + 9 + 10 + 11 + 12 + 13 + 15 + 16 = 104 R2: 2 + 3 + 6 + 7 + 14 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 172 n1 n1 n2 1 111112 1 1124 264 R: 132 R 2 2 2 2 σR: n n n n 1 11121112 1 1112 24 3168 1 2 1 2 264 16.248 R 12 12 12 12 R 104 132 28 Test statistic: z R 1.723280874 1.723 R 16.248 16.248 This is a two-tailed test and the critical values (from Table A-2) are: ±1.645 Since the test statistic is in the critical range, we reject the null hypothesis and conclude that samples do not have equal medians. Check: OK
Section 13-5 – Basic Skills and Concepts
1) What major advantage does the Kruskal-Wallis test have over the one-way ANOVA test? The samples used for the Kruskal-Wallis test do not have to come from populations with a certain distribution. Check: OK
3) Refer to Table 13-2 and identify the efficiency of the Kruskal-Wallis test. What does that value tell us about the test? The efficiency rating for the Kruskal-Wallis test is 0.95. This means that Kruskal- Wallis test would require 100 observations where as the parametric counterpart, ANOVA, would only require 95. Overall, this nonparametric test is pretty efficient as compared to ANOVA. Check: OK
5) Data were obtained from car crash experiments conducted by the National Transportation Safety Administration. New cars were purchases and crashed into a fixed barrier at 35 MPH, and measurements were recorded for the dummy in the driver’s seat. Use the sample data listed below to test for difference in head injury measurements (in hic) among the four weight categories. Is there sufficient evidence to conclude that head injury measures for the four car weight categories are not all the same? Do the data suggest that heavier cars are safer in a crash? Subcompact 681 428 917 898 420
RANK 16 5 20 19 4 R1 = 64 Compact 643 655 442 514 525
RANK 13 14 6 9 10.5 R2 = 52.5 Midsize 469 727 525 454 259
RANK 8 18 10.5 7 1 R3 = 44.5 Full-size 384 656 602 687 360
RANK 3 15 12 17 2 R4 = 49
H0: The measurements from the four weight classes have the same medians. H1: The four population medians are not equal. N: 20 k = 4 R1: 64 R2: 52.5 R3: 44.5 R4: 49 2 2 2 2 12 R1 R2 R3 R4 H 3N 1 NN 1 n1 n2 n3 n4 12 642 52.52 44.52 492 320 1 H: 2020 1 5 5 5 5 12 819.2 551.25 396.05 480.2 321 2021 0.02857142862246.7 63 64.19142864 63 1.19142864 1.191 The critical value (with 3 degrees of freedom) is: 7.815 Since the test statistic is NOT in the critical range, we fail to reject the null hypothesis and conclude that samples do have equal medians. Heavier cars are not necessarily safer that other cars. Check: OK
Section 13-6 – Basic Skills and Concepts
1) What major advantages does the rank correlation method of this section have over the linear correlation method of Section 10-2? The first advantage is that samples used for the rank correlation test do not have to come from populations with a certain distribution. In addition, the rank correlation method may also detect non-linear relationships. Check: OK
3) We represent the rank correlation coefficient test statistic the notation rs, and the corresponding population parameter is represented as ps. Why is the subscript s used? Does the subscript s represent the same standard deviation s introduced in Section 3-3? The s is in honor of Charles Spearman. The s notation has nothing to do with standard deviation. Check: OK
5) Sketch a scatter diagram, find the value of rs, and determine whether there appears to be a correlation between x and y. X 2 4 1 5 3 RANK 3.5 7.5 1.5 9.5 5.5 Y 2 4 1 5 3 RANK 3.5 7.5 1.5 9.5 5.5 Difference 0 0 0 0 0 The scatter plot looks like this: Scatter Plot of X and Y
5.00
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y 3.00
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1.00 2.00 3.00 4.00 5.00 x
6 d 2 602 ... 02 60 rs: r 1 1 1 ... 1 0 1 2 nn 2 1 552 1 5(24)
Because rs = 1, there is perfect correlation between x and y. More specifically, it is a perfect, positive linear correlation. Check: OK
Section 13-7 – Basic Skills and Concepts
1) A pollster conducts a survey by filling out a page of responses from each subject he interviews. He then shuffles the pages before submitting them for analysis. Can the run test for randomness be used to determine whether his survey subjects were selected from a random sequence. No. The shuffling process happened after the subjects were already selected. There is no information about how the participants were initially selected. Check: OK
5) Use the given sequence to determine the values of n1, n2, the number of runs G, and the critical values from Table A-10, and use those results to determine whether the sequence appears to be random. Y N N Y N N Y N N Y N N n1: The total number of yes’s is: 4. n2: The total number of no’s is: 8. G: The total number of runs: 8. The critical values are 3 and 10. The test statistic is G since both n1 and n2 are less than or equal to 20. Since G is not less than or equal to 3 or greater than or equal to 10, we do not reject randomness. Check: OK 10) Use the runs test of this section to determine whether the given sequence is random. Use a significance level of = 0.05. Listed below are gold medal winners in the men’s 400-meter dash, where U represents the United States and O represents any other country. Does it appear that the countries of the winners are in a random sequence? U U U U O U O O U U U O O U U U U U O O U U U U U U n1: The total number for the U.S. is: 19. n2: The total number for other countries: 7. G: The total number of runs: 9. The critical values are 6 and 16. The test statistic is G since both n1 and n2 are less than or equal to 20. Since G is not less than or equal to 6 or greater than or equal to 16, we do not reject randomness. Check: OK
Unit 13 – Review Exercises
1) Use a 0.05 significance level. Use the appropriate nonparametric test from this chapter. The table lists matched pairs of time obtained from a random sample of children who were given blocks and instructed to build a tower as tall as possible. This procedure is used to measure intelligence in children. Use the sign test and 0.05 significance level to test the claim that there is no difference between the times of the first and second trials. Child A B C D E F G H I J K L M N O 1st 30 19 19 23 29 178 42 20 12 39 14 81 17 31 52 2nd 30 6 14 8 14 52 14 22 17 8 11 30 14 17 15 Sign 0 + + + + + + – – + + + + + + Positive signs: 12 Negative signs: 2 Ties: 1
The null hypothesis is: H0: There is no difference.
The alternative hypothesis is: H1: There is a difference (the median of differences 0) The significance level is: = 0.05 Use the nonparametric sign test. The test statistic is: x = 2 (the smaller of the two counts). N = 14 since ties are discarded. This is a two-tailed test and the critical value (from Table A-7) is: 2. Since the test statistic is less than or equal to 2, we reject the null hypothesis and conclude that there is a difference in the times from the first trial to the second. Check: OK
3) Use a 0.05 significance level. Use the appropriate nonparametric test from this chapter. U.S. News and World Report magazine ranked business schools and law schools, and the ranks in the table below are based on those results. Is there a correlation between the business school rankings and the law school rankings?
Business 1 6 2 5 4 3 10 8 9 7 RANK 1.5 11.5 3.5 9.5 7.5 5.5 19.5 15.5 17.5 13.5 Law 2 1 3 4 5 6 9 10 7 8 RANK 3.5 1.5 5.5 7.5 9.5 11.5 17.5 19.5 13.5 15.5 Difference –2 10 –2 2 –2 –6 2 –4 2 –1 The scatter plot looks like this:
Scatter Plot of School Rankings
Business and Law Schools
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R Sq Linear = 0.503 2
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0 2 4 6 8 10 law
rs (since there are ties): x 55 y 55 x 2 385 y 2 385 xy 361
n xy x y 10 361 5555 r2 2 2 2 2 n x 2 x n y 2 y 10 385 55 10 385 55 3610 3025 585 0.70909090909 0.709 3850 3025 3850 3025 825 The critical value is 0.648 Because rs = 0.709 and is in the critical region, we can conclude that a correlation does exist between the rankings for the business and law schools at each university. Check: OK. I had to go back and use the procedure for ties since ties occurred in the ranking. Once I made this realization, the rest of the process went fine. 5) Use a 0.05 significance level. Use the appropriate nonparametric test from this chapter. The Axipon accounting firm claims that hiring is done without gender bias. Among the 40 new employees hired, 15 are women. Job applicants are about half men and half women, who are all qualified. Is there sufficient evidence to charge bias in favor of men? The null hypothesis is: H0: p = 0.5.
The alternative hypothesis is: H1: p 0.5 The significance level is: = 0.05 Use the nonparametric sign test. Women are positive (+) signs. Men are minus (–) signs. There are 15 women and 25 men. The test statistic is: n 40 x 0.5 15 0.5 2 2 15 0.5 20 4.5 z 1.423 n 40 6.32455532 3.16227766 2 2 2 This is a two-tailed test and the critical value (from Table A-2) is: ±1.645. Since the test statistic is NOT in the critical region, we fail to reject the null hypothesis and conclude that there is not bias in the hiring process at Axipon. Check: OK
Unit 13 – Cumulative Review Exercises
1) Use the data in table for the before scores. 460 470 ... 620 610 4760 Mean: 528.888888 528.9 9 9 Median: 460 470 490 490 510 510 600 610 620
The median is: 510 SD: x 2 4602 4702 ... 6202 6102 2549400 2 x (460 470 ... 620 610)2 47602 22657600 2 n x 2 x 9(2549400) 22657600 287000 s nn 1 99 1 72 3986.1111111 63.13565642 63.136
Variance: s 2 63.135656422 3986.111111111 3986.111
Range: 620 460 160 Check: OK